Question

The 14 term of an AP is twice its 8 term. If its 6 term is - 8. then find the sum of its first 20 terms.​

Answer 1

ATQ,

The 14th term of an AP is twice it's 8th term.

We know that,

a14 = a + 13d

Similarly

a8 = a + 7d

Since 14th term is twice it's 8th term,

➡ 2a8 = a14

➡ 2(a + 7d) = a + 13d

➡ 2a + 14d = a + 13d

➡ 2a - a = 13d - 14d

➡ a = -1d

➡ a + d = 0 ------(i)

Now, Given that it's 6th term is -8

➡ a + 5d = -8 ------(ii)

By solving equation (i) and equation (ii), we get

➡ d = 2

Putting value of d = -2 in equation (i)

➡ a + (-2) = 0

➡ a - 2 = 0

➡ a = 2

Hence the sum of first 20 terms of the AP = n/2 [2a + (n - 1)d]

= 20/2 [(2 × 2) + (19 × -2)]

= 10(4 - 38)

= 10 × -34

= -340 Final answer!

Answer 2

Answer:

The 14 term of an AP is twice is 8 term .

Formula :

= a + (n - 1)d

= a + (14 - 1) d = a + 13d

= a + (8 - 1) d = a + 7d

= a + (6 - 1) d = a + 5d

= a + 13d = 2(a + 7d)

= a + d = 0 .....(i)

and

a + 5d = -8 ...(ii)

Solving (i) and (ii)

a = 2

d = -2

The sum of its first = 20 terms.

= 20/2 [2 × 2 + (20 - 1) (-2)]

= - 340

= - 340.